Question

Identify A to in the following sequence of reaction.
$Na+O_2\to \left(A\right)\stackrel{HCl}{\to }\left(B\right)\underset{H^{+}/K_{2}C{r}_2{O}_7}{\overset{ether}{\to }}C$

$C$ is forms blue colour in ethereal solution.

• A. A: $N{a}_2{O}_2$, B : $H_{2}O$, C : $Cr{O}_3$
• B. A: $Na{O}_2$, B : $H_2{O}_2$, C : $C{r}_2{O}_7$
• C. A: $N{a}_2{O}_2$, B : $H_2{O}_2$, C : $Cr{O}_3$
• D. A: $NaO$, B : $H_{2}O$, C : $Cr{O}_3$

Answer 1

The correct option is C A: $N{a}_2{O}_2$, B : $H_2{O}_2$, C : $Cr{O}_3$

$2Na+O_2\to N{a}_2{O}_2$
$N{a}_2{O}_2+2HCl\to H_2{O}_2+2NaCl$
$K_{2}C{r}_2{O}_7+H_{2}S{O}_4+H_2{O}_2\to 2Cr{O}_3+K_{2}S{O}_4+2H_{2}O$
So, $A$: $N{a}_2{O}_2$, $B$ : $H_2{O}_2$, $C$ : $Cr{O}_3$.
Hence, option $C$ is correct.